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Q1) What is the rightmost non-zero digit of 20!? a) 2 b) 3 c) 5 d) 6 e) 8

Q2) What is the product of the factors of 432? a) 2^90 b) 3^78 c) 2^40 * 3^30 d) 2^25 * 3^75 e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

The easiest way to solve Q2 is as follows:

There is a formula to find the product of factors of a number 'n'. => (Prime factorization of n) ^ f/2 , where f is the total no. of factors.

Coming back to the quqestion, prime factorization of 432 = 2^4 * 3^3 Hence, total no.of factors = (4+1)*(3+1) = 5*4 = 20 The product of all factors of 432 = (2^4 * 3^3)^(20/2) = (2^4 * 3^3)^10 = 2^40 * 3^30 = C

Set 1 : {2,12} Set 2 : {5,15} Set 3 : {10,20} Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19}

This reduces to (canceling 2s&5s in set 1 and set 2 + canceling 10s in set 3) :

Set 1 : {1,6} --> Last digit 6 Set 2 : {1,3} --> Last digit 3 Set 3 : {1,2} --> Last digit 2 Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19} --> Last digit 8*8=4

So overall last digit = 6*3*2*4 = 4

Hence last non-zero digit of 20! = 4

Can you please explain how you calculating the last digit for Set 4 ?

Set 4 is everything else {1,3,4,6,7,8,9,11,13,14,16,17,18,19, 21,23,24,26,27,28,29.........81,83,84,86,87,88,89} notice the repeating pattern is set 4 - 1,3,4,6,7,8,9,1,3,4,6,7,8,9,........ and there are 9 such patterns (1,3,4,6,7,8,9) (11,13,14,16,17,18,19).....(81,83,84,86,87,88,89)

Multiply the digits in the patter 1*3*4*6*7*8*9 - the units digit is 8

Quote:

The "cyclicity" of 8 in a product is \(8^1=8\) \(8^2=64\) \(8^3=512\) \(8^4=4096\) \(8^5=32768\) and so on, the pattern is 8,4,2,6....

And as the pattern (1,3,4,6,7,8,9) repeats 9 times - so the units digit will be 8
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There is a brute force way to solve this. The principle is the same, you have to remove all powers of 5 and a corresponding number of powers of 2 from the product. The product of the rest will give you the non-zero digit.

First of all what I'd do is split the numbers out into four sets. Set 1 is {2,12,22, ... ,82}. Set 2 is {5,15,25,...,85}. Set 3 is {10,20,30,...90} and Set 4 is everything else {1,3,4,6,7,8,9, 11,13,14,16,17,18,19, .... ,81,83,84,86,87,88,89}

Set 1,2,3 have 9 elements each and have all the powers of 5 within them. Set 4 has the remaining 7x9=63 elements

Lets process Set 3 first, I can take out all powers of 10, to get {1,2,3,4,5,6,7,8,9} Then I take out the 5 and a corresponding 2 to get {1,1,3,4,1,6,7,8,9}

Now lets process Set 1 & 2, For every 5 I take out of set 2, I will take out a 2 from set 1 : So set 1 becomes : {1,6,11,16,....,36,41} and set 2 : {1,3,5,7,9,11,13,15,17}

We are almost done now, the only powers of 5 left are in set 2 which I take out and against that I take out 4 in set 3 to leave me with with :

Set 1 : {1,6,.....,36,41} : Last digit is 6 Set 2 : {1,3,1,7,9,11,13,3,17} : Last digit is 7 Set 3 : {1,1,3,1,1,6,7,8,9} : Last digit 2 Set 4 : Which has a pattern to it and easy to show last digit is 8 (It has subsets ending in 1,3,4,6,7,8,9 each of which has last digit 8 and 9 such subsets. The "cyclicity" of 8 in a product is 8,4,2,6,repeat so with 9 subsets the answer will be 8) Multiplying those out, the last digit of 90! is 2

The approach is brute force, but once you get a hang of it, it doesnt take very long to calculate it out. I would be very surprised if this is a GMAT level question, unless of course there is a trick to doing this which has completely evaded me

Hi,

I dont understand set 3.. Its ok you took out 2 and 5 , so that they become 1 and 1 respectively. But why 4 in set 3 becomes 1 ..Will you please explain?
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Kabilan.K Kudos is a boost to participate actively and contribute more to the forum

a) 2^90 b) 3^78 c) 2^40 * 3^30 d) 2^25 * 3^75 e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

Thats a good & tough question:

a) 2^90 ......... cannot be because it doesnot have any multiple/power of 3. b) 3^78......... cannot be because it doesnot have any multiple/power of 2. c) 2^40 * 3^30......... can be because it has multiples/power of 2 & 3. d) 2^25 * 3^75 ......... can be because it has multiples/power of 2 & 3. e) 2^5 * 3*10 * (2^25 * 3^30 - 1) probably not.

Lets work on c and d. 432 = (2^4) (3^3) Factor of 432 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 108, 144, 216 and 432.

Product of 432 = 2x3x4x6x8x9x12x16x18x24x27x36x48x54x72x108x144x216x432 = 2x3x 2^2 x 2x3 x 2^3 x 3^3 x 2^2x3 x 2^4 x 2x3^2 x 2^3x3 x 3^3 x 2^2x3^2 x 2^4x3 x 2x3^3 x 2^3x3^2 x 2^2x3^3 x 2^4x3^2 x 2^3x3^3 x 2^4x3^3 = 2^41 x 3^31

Hmmm ........ I got 2^41 x 3^31 instead of 2^40 x 3^30.

Answer should be C but how it is 2^40 x 3^30? I must have made a mistake somewhere....

Can you find it for me? Thanx....

2x 3x 2^2 x 2x3 x 2^3 x 3^3(p) x 2^2x3 x 2^4 x 2x3^2 x 2^3x3 x 3^3 x 2^2x3^2 x 2^4x3 x 2x3^3 x 2^3x3^2 x 2^2x3^3 x 2^4x3^2 x 2^3x3^3 x 2^4x3^3

3^3(p) x For 9, you have written 3^3 instead of 3^2. But thts all I could find out.

I counted 40 2s, so that is ok...you miscounted...
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Re: What is the rightmost non-zero digit of 20!? [#permalink]

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27 Jan 2014, 09:06

Question 1:

For the first digit after the 0s, we just multiply the unit digits excluding 0,2,5s \(1*3*4*6*7*8*9 = 36288\) (if you multiply step by step, drop the whole number and only use the unit digits of your solutions!)

This needs to be done twice, because the unit digits appear between 0 and 10 as well as between 10 and 20.

So you multiply the unit digit from 36288 with 36288: \(8*8 = 64\) --> First digit after the 0s is 4.

If you want to compute this with 90!, you've got to calculate \(8^9\) (or step by step with the unit digit of your solutions

\(64^4*8 -> 4^4 * 8 -> 6^2 * 8 -> 6*8 = 48\) Thus the first digits (after the 21) 0s is 8. (back-checked with wolframalpha)

Question 2: (did not use timer, but guess between 1min and 1:30)

\(432 = 2^4 * 3^3\)

Attachment:

Screen Shot 2014-01-27 at 18.03.33.png [ 11.21 KiB | Viewed 1323 times ]

According to the table, the solution is \(2^40 * 3^30\)
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Thank You = 1 Kudos B.Sc., International Production Engineering and Management M.Sc. mult., European Master in Management Candidate

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